(a) $T_1=a=-8$
$\frac{T_7}{T_9}=\frac{5}{8}$
$\frac{a+6d}{a+8d}=\frac{5}{8}⟹\frac{-8+6d}{-8+8d}=\frac{5}{8}$
$-8+6d=\frac{5}{8}(-8+8d)⟹-8+6d=-5+5d$
$6d-5d=-5+8$
$d=3$
(b) Let :
the cost price $C{P}_1$ of a basket of pawpaw be w;
the cost price $C{P}_2$ of a basket of mangoes be m.
Then $30w+100m=N2,450.....(1)$
Also, let :
The selling price of the baskets of pawpaw be $S{P}_1$;
the selling price of the baskets of mangoes be $S{P}_2$.
Then $S{P}_1+S{P}_2=N(2,450+855)=N3,305......(2)$
%age profit on the sale of pawpaw = $\frac{S{P}_1-C{P}_1}{C{P}_1}\times 100$
i.e. $40$
$⟹40\times 3w=(S{P}_1-30w)\times 10$
$12w=S{P}_1-30w$
$S{P}_1=12w+30w=42w$
%age profit on the sale of mangoes = $\frac{S{P}_2-C{P}_2}{C{P}_2}\times 100$
i.e. $30$
$⟹30m=S{P}_2-100m$
$S{P}_2=30m+100m=130m$
But $S{P}_1+S{P}_2=N3,305$
$⟹42w+130m=3305....(2)$
From (1), $m=\frac{2450-30w}{100}$
$m=\frac{245-3w}{10}$
Substitute $m=\frac{245-3w}{10}$ in (3)
$42w+130(\frac{245-3w}{10})=3305$
$42w+3,185-39w=3305$
$3w=3305-3185=120$
$w=N40.00$
(i) Hence, the cost price of a basket of pawpaw = N40.00.
(ii) From (2),
$S{P}_1+S{P}_2=3305$
$42w+130m=3305⟹42\times 40+130m=3305$
$1680+130m=3305$
$130m=S{P}_2=3305-1680$
= $N1,625.00$
Hence, the selling point of 100 baskets of mangoes = N1,625.00.